3.1: Radical Equations and Extraneous Solutions | Algebra and Trigonometry

How it works

Isolate a radical before raising both sides to a power. Squaring can introduce solutions that do not satisfy the original equation. Record the original real-domain restrictions and check every candidate in the equation before reporting it.

\sqrt{A}=B\Rightarrow A=B^2\text{ and }B\ge0

Worked examples

Example 1

Solve $\displaystyle \sqrt{x+5}=x-1$ .

The right side requires $\displaystyle x\ge1$ . Squaring gives $\displaystyle x+5=x^2-2x+1$ , so $\displaystyle (x-4)(x+1)=0$ . Only $\displaystyle x=4$ checks.

Example 2

Solve $\displaystyle \sqrt{2x+3}=5$ .

Square to get $\displaystyle 2x+3=25$ , so $\displaystyle x=11$ . Substitution gives $\displaystyle \sqrt{25}=5$ .

Problems

Problem 3.1.1

Solve $\displaystyle \sqrt{x+1}=3$ .

Show worked solution

$\displaystyle x+1=9$ , hence $\displaystyle x=8$ , which checks.

Problem 3.1.2

Solve $\displaystyle \sqrt{x}=x-2$ .

Show worked solution

Require $\displaystyle x\ge2$ . Squaring gives $\displaystyle x^2-5x+4=0$ , so candidates are $\displaystyle 1$ and $\displaystyle 4$ . Only $\displaystyle x=4$ satisfies the original.

Problem 3.1.3

Use the graphs of $\displaystyle y=\sqrt{x+2}$ and $\displaystyle y=x$ to solve $\displaystyle \sqrt{x+2}=x$ .

The curves y = square root of (x + 2) and y = x meet at positive x. — A valid solution must satisfy the original radical equation.

Show worked solution

Squaring gives $\displaystyle (x-2)(x+1)=0$ . The radical is nonnegative, so $\displaystyle x=-1$ fails. The valid intersection is $\displaystyle (2,2)$ .

The curves y = square root of (x + 2) and y = x meet at positive x. The intersection is (2, 2); the squared-equation candidate −1 is extraneous. — The intersection is (2, 2); the squared-equation candidate −1 is extraneous.

Lesson Problems PDF Solutions PDF Unit MCQ

How it works

Worked examples

Example 1

Example 2

Problems

Problem 3.1.1

Problem 3.1.2

Problem 3.1.3

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